Question

For a correlation matrix of $n$ random variables:

  1. What is the sum of eigenvalues?
  2. What is a lower bound for the sum of eigenvalues of the inverse of a nonsingular $n\times n$ correlation matrix?
Solution

Sum of eigenvalues.

For any square matrix, the sum of eigenvalues equals the trace.
A correlation matrix has all diagonal entries equal to $1$, so $$ \sum_{i=1}^n \lambda_i = \operatorname{tr}(C) = n. $$

Solution

Lower bound for the eigenvalue sum of the inverse.

Let $\lambda_1,\dots,\lambda_n$ be the eigenvalues of a nonsingular correlation matrix.
We know: $$ \sum_{i=1}^n \lambda_i = n. $$

Apply Cauchy–Schwarz to the vectors
$\left(\sqrt{\lambda_1},\dots,\sqrt{\lambda_n}\right)$ and
$\left(\frac{1}{\sqrt{\lambda_1}},\dots,\frac{1}{\sqrt{\lambda_n}}\right)$: $$ \left( \sum_{i=1}^n \sqrt{\lambda_i}\cdot \frac{1}{\sqrt{\lambda_i}} \right)^2 \le \left( \sum_{i=1}^n \lambda_i \right) \left( \sum_{i=1}^n \frac{1}{\lambda_i} \right). $$

Thus, $$ n^2 \le n \sum_{i=1}^n \frac{1}{\lambda_i}. $$

Canceling $n$ gives: $$ \sum_{i=1}^n \frac{1}{\lambda_i} \ge n. $$

Since the eigenvalues of $C^{-1}$ are $\frac{1}{\lambda_i}$, the lower bound is $$ \sum_{i=1}^n \lambda_i(C^{-1}) \ge n. $$